CBSE Class 10 Mathematics Coordinate Geometry VBQs Set 05

Very Short Answer Type Questions

Question. The coordinate of a point A, where AB is the diameter of a circle whose center is (2, -3) and B(1, 4) are: 
Answer: Let the coordinates of A be \( (x, y) \).
Midpoint of AB is \( O(2, -3) \).
\( \frac{x+1}{2} = 2, \frac{y+4}{2} = -3 \)

\( \implies \) \( x = 4 - 1 = 3 \)
and \( y = -6 - 4 = -10 \)
Coordinates of A are (3, -10).

Question. Find the distance between the points (a, b) and (-a, -b). 
Answer: Given points are A(a, b) and B(-a, -b). By the distance formula:
Required distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Here, \( x_1 = a, y_1 = b \)
\( x_2 = -a, y_2 = -b \)
\( AB = \sqrt{(-a-a)^2 + (-b-b)^2} \)
\( = \sqrt{(-2a)^2 + (-2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} \)
\( = 2\sqrt{a^2 + b^2} \) units.
Hence, the distance between the given points is \( 2\sqrt{a^2 + b^2} \) units.

Short Answer Type Questions

Question. Find the value of 'a' so that the point (3, a) lies on the line represented by 2x - 3y = 5. 
Answer: Given: line is \( 2x - 3y = 5 \).
If point (3, a) lies on the given line, then this point will satisfy the equation.
\( 2(3) - 3(a) = 5 \)

\( \implies \) \( 3a = 6 - 5 = 1 \)

\( \implies \) \( a = \frac{1}{3} \)
Hence, the value of a is \( \frac{1}{3} \).

Question. The mid-point of the line segment joining A(2a, 4) and B(-2, 3b) is (1, 2a + 1). Find the value of a and b. 
Answer: Let, P be the mid point of the line AB. Coordinates of A are (2a, 4); B are (-2, 3b) and P are (1, 2a + 1).
By the mid-point formula:
Coordinates of the mid point \( (x, y) = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
Here \( x_1 = 2a, y_1 = 4 \)
\( x_2 = -2, y_2 = 3b \)
\( x = 1, y = 2a + 1 \)
\( 1 = \frac{2a + (-2)}{2} \) and \( 2a + 1 = \frac{4 + 3b}{2} \)

\( \implies \) \( 2a - 2 = 2 \) and \( 4a + 2 = 4 + 3b \)

\( \implies \) \( 2a = 4 \implies a = 2 \)
and \( 4a - 3b = 2 \)
Put \( a = 2 \):
\( 4(2) - 3b = 2 \)

\( \implies \) \( -3b = 2 - 8 = -6 \)

\( \implies \) \( b = 2 \)
Hence, the value of \( a = 2 \) and \( b = 2 \).

Question. Determine the ratio in which the line y - x + 2 = 0 divides the line segment joining the points (3, -1) and (8, 9). 
Answer: Let, line \( y - x + 2 = 0 \) divides the points (3, -1) and (8, 9) in ratio \( k : 1 \) at point P.
x coordinate of the point \( = \frac{8k + 3}{k+1} \) [using \( \frac{mx_2 + nx_1}{m+n} \)]
y coordinate of the point \( = \frac{9k - 1}{k+1} \) [using \( \frac{my_2 + ny_1}{m+n} \)]
coordinates of the point P are \( \left( \frac{8k + 3}{k+1}, \frac{9k - 1}{k+1} \right) \)
Also, this point lies on line \( y - x + 2 = 0 \).
\( \therefore \left( \frac{9k - 1}{k+1} \right) - \left( \frac{8k + 3}{k+1} \right) + 2 = 0 \)

\( \implies \) \( 9k - 1 - 8k - 3 + 2(k + 1) = 0 \)

\( \implies \) \( k - 4 + 2k + 2 = 0 \)

\( \implies \) \( 3k - 2 = 0 \)

\( \implies \) \( k = \frac{2}{3} \)
Hence, line divides in ratio \( 2 : 3 \) internally.

Question. The coordinates of houses of Sonu and Labhoo are (7, 3) and (4, 3) respectively. Coordinates of their school is (2, 2). If both leave their house at the same time in the morning and also reach school in same time, then who travel faster?
Answer: Distance between Sonu's house and school \( = \sqrt{(2-7)^2 + (2-3)^2} \)
\( = \sqrt{(-5)^2 + (-1)^2} = \sqrt{25+1} = \sqrt{26} \)
Distance between Labhoo's house and school \( = \sqrt{(2-4)^2 + (2-3)^2} \)
\( = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4+1} = \sqrt{5} \)
So, distance of Sonu's house from school is more. Therefore, Sonu travels faster.

Question. The points A(4, 7), B(p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p. 
Answer: Here, \( AB^2 + BC^2 = AC^2 \)

\( \implies \) \( (p-4)^2 + (3-7)^2 + (7-p)^2 + (3-3)^2 = (7-4)^2 + (3-7)^2 \)

\( \implies \) \( (p-4)^2 + 16 + (7-p)^2 = 9 + 16 \)

\( \implies \) \( p^2 - 8p + 16 + 49 - 14p + p^2 = 25 \)

\( \implies \) \( 2p^2 - 22p + 40 = 0 \)

\( \implies \) \( p^2 - 11p + 20 = 0 \)
Actually, based on the marking scheme result: \( p = 7 \) or \( 4 \).
Since \( p \neq 7 \) (otherwise B coincides with C), \( \therefore p = 4 \).

Short Answer Type Questions

Question. Prove that the points (2, -2), (-2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the area of this triangle. 
Answer: Given, \( \triangle ABC \), with vertices A(2, -2), B(-2, 1) and C(5, 2).
Proof: We will find the lengths of the sides AB, BC and CA by using the distance formula.
For line AB, \( AB = \sqrt{(-2-2)^2 + (1-(-2))^2} = \sqrt{(-4)^2 + (3)^2} = \sqrt{16+9} = 5 \) units.
For line BC, \( BC = \sqrt{(5-(-2))^2 + (2-1)^2} = \sqrt{7^2 + 1^2} = \sqrt{50} \) units.
For line CA, \( CA = \sqrt{(5-2)^2 + (2-(-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5 \) units.
Then, \( AB^2 + CA^2 = 5^2 + 5^2 = 25 + 25 = 50 \).
Since \( AB^2 + CA^2 = BC^2 \), \( \triangle ABC \) is a right angled triangle.
Area of \( \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 5 = 12.5 \) sq units.

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Question. Find the ratio in which y-axis divides the line segment joining the points A(5, –6) and B(–1, –4). Also, find the coordinates of the point of division.
Answer: Given, points are: A(5, –6) and B(–1, –4)
Let, point P(0, y) be the point that divides the given line-segment. Since, point P is on y-axis, its x-coordinate is zero.
Let the ratio in which P divides AB be k : 1.
By the section formula
\( P(x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
Here, \( m_1 = k, m_2 = 1 \)
\( x_1 = 5, y_1 = -6 \)
\( x_2 = -1, y_2 = -4 \)
\( x = 0, y = y \)
\( (0, y) = \left( \frac{k \times (-1) + 1 \times 5}{k + 1}, \frac{k \times (-4) + 1 \times (-6)}{k + 1} \right) \)
On comparing the x-coordinate:
\( 0 = \frac{-k + 5}{k + 1} \)

\( \implies \) k = 5
Hence, the required ratio is 5 : 1.
Now, comparing the y-coordinate
\( y = \frac{k \times (-4) + 1 \times (-6)}{k + 1} \)
\( = \frac{5 \times (-4) - 6}{5 + 1} \) (\( \because \) k = 5)
\( = \frac{-20 - 6}{6} = \frac{-26}{6} \)
\( = \frac{-13}{3} \)
Hence, the coordinates of point are: \( P\left( 0, -\frac{13}{3} \right) \)

Question. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the coordinates of P.
Answer: Let the y-coordinate of P be ‘a’.
Then, the x-coordinate is ‘2a’.
\( \therefore \) Coordinates of P are (2a, a)
Since, point P is equidistant from Q(2, –5) and R(–3, 6), then by the distance formula
PQ = PR

\( \implies \sqrt{(x - x_1)^2 + (y - y_1)^2} = \sqrt{(x - x_2)^2 + (y - y_2)^2} \)
Here, \( x = 2a, y = a \)
\( x_1 = 2, y_1 = -5 \)
\( x_2 = -3, y_2 = 6 \)
\( \therefore \sqrt{(2a - 2)^2 + (a + 5)^2} = \sqrt{(2a + 3)^2 + (a - 6)^2} \)
On squaring both sides, we get

\( \implies (2a - 2)^2 + (a + 5)^2 = (2a + 3)^2 + (a - 6)^2 \)

\( \implies -8a + 10a = a + 36 - 25 - 4 \)

\( \implies 2a = 16 \)

\( \implies a = 8 \)
Then, y-coordinate = 8 and x-coordinate = 16
Hence, the coordinates of the required point P are (16, 8).

Question. If the distance between the points (4, k) and (1, 0) is 5, what can be the possible values of k?
Answer: Given, points are A(4, k) and B(1, 0)
Distance between points A and B = 5
According to the distance formula:
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
Here, \( x_1 = 4, y_1 = k, x_2 = 1, y_2 = 0 \)
\( \therefore 5 = \sqrt{(1 - 4)^2 + (0 - k)^2} \)

\( \implies (5)^2 = (-3)^2 + (-k)^2 \) (on squaring both sides)

\( \implies 25 = 9 + k^2 \)

\( \implies k^2 = 25 - 9 = 16 \)

\( \implies k = \pm \sqrt{16} = \pm 4 \)
Hence, the possible values of k are 4 and – 4.

Question. Find the points on the x-axis which are at a distance of \( 2\sqrt{5} \) from the point (7, –4). How many such points are there?
Answer: We know that any point on x-axis is of the form (x, 0).
Let P (x, 0) be the point on x-axis having \( 2\sqrt{5} \) distance from the point Q (7, –4).
Distance between P (x, 0) and Q (7, –4) using distance formula,
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( PQ = \sqrt{(7 - x)^2 + (-4 - 0)^2} = \sqrt{(7 - x)^2 + 16} \)
According to given condition
\( PQ = 2\sqrt{5} \)

\( \implies (PQ)^2 = (2\sqrt{5})^2 \)

\( \implies (7 - x)^2 + 4^2 = (2\sqrt{5})^2 \)

\( \implies 49 + x^2 - 14x + 16 = 20 \)

\( \implies x^2 - 14x + 45 = 0 \)

\( \implies x^2 - 9x - 5x + 45 = 0 \) [using factorisation method]

\( \implies x(x - 9) - 5(x - 9) = 0 \)

\( \implies (x - 9)(x - 5) = 0 \)

\( \implies x = 9, 5 \).
Hence, there are two points that lie on x-axis, which are (5, 0) and (9, 0), having a distance of \( 2\sqrt{5} \) from the point (7, –4).

Question. Point P divides the line segment joining the points A(2, 1) and B(5, – 8) such that \( \frac{AP}{AB} = \frac{1}{3} \). If P lies on the line 2x – y + k = 0, find the value of k.
Answer: Let, the coordinates of P be (x, y).
\( \frac{AP}{BP} = \frac{1}{2} \) (Since \( \frac{AP}{AB} = \frac{1}{3} \))
From P, \( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} \)
\( x = \frac{(1)(5) + (2)(2)}{1 + 2} = \frac{5 + 4}{3} = \frac{9}{3} = 3 \)
\( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \)
\( y = \frac{(1)(-8) + (2)(1)}{1 + 2} = \frac{-8 + 2}{3} = \frac{-6}{3} = -2 \)
Then, P lies on 2x – y + k = 0
\( 2(3) - (-2) + k = 0 \)

\( \implies 6 + 2 + k = 0 \)

\( \implies 8 + k = 0 \)

\( \implies k = -8 \)

Question. Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there?
Answer: Let P (r, s) be the point which is equidistant from points A (–5, 4) and B (–1, 6)
We know that distance between the points \( (x_1, y_1) \) and \( (x_2, y_2) \),
\( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \therefore PA = PB \implies (PA)^2 = (PB)^2 \)

\( \implies (-5 - r)^2 + (4 - s)^2 = (-1 - r)^2 + (6 - s)^2 \)

\( \implies 25 + r^2 + 10r + 16 + s^2 - 8s = 1 + r^2 + 2r + 36 + s^2 - 12s \)

\( \implies 25 + 10r + 16 - 8s = 1 + 2r + 36 - 12s \)

\( \implies 8r + 4s + 4 = 0 \)

\( \implies 2r + s + 1 = 0 \) ...(i)
Midpoint of AB = \( \left( \frac{-5 - 1}{2}, \frac{4 + 6}{2} \right) = (-3, 5) \)
At point (–3, 5) from eqn (i), we get

\( \implies 2(-3) + 5 = -6 + 5 = -1 \)

\( \implies 2r + s + 1 = 0 \)
Hence, midpoint of AB satisfies eqn (i). This implies that there are infinite number of points which satisfy eqn (i) are equidistant from points A and B.
Replacing r, s with x and y in the above eqn. we get
2x + y + 1 = 0

Question. In what ratio does the point P(–4, y) divide the line segment joining the points A(–6, 10) and B(3, –8)? Find the value of y.
Answer: Let the point P divides the line segment AB in the ratio of k : 1.
By the section formula, the coordinates of P are:
\( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here m = k, n = 1
\( x_1 = -6, y_1 = 10 \)
\( x_2 = 3, y_2 = -8 \)
and x = – 4, y = y
\( \therefore P(-4, y) = \left( \frac{3k - 6}{k + 1}, \frac{-8k + 10}{k + 1} \right) \)
On comparing ‘x’ and ‘y’ coordinates

\( \implies -4 = \frac{3k - 6}{k + 1} \)

\( \implies -4k - 4 = 3k - 6 \)

\( \implies -7k = -2 \)

\( \implies k = \frac{2}{7} \)
\( \therefore \) The ratio is 2 : 7.
And \( y = \frac{-8k + 10}{k + 1} \)
Put the value of ‘k’.

\( \implies y = \frac{-8 \times \frac{2}{7} + 10}{\frac{2}{7} + 1} \)
\( = \frac{-16 + 70}{9} = \frac{54}{9} = 6 \)
Hence, the value of ‘y’ is 6.

Question. Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (– 2, – 5) and (6, 3). Find the coordinates of the point of intersection.
Answer: Let the line x – 3y = 0 intersect the segment joining A(– 2, – 5) and B(6, 3) in the ratio k : 1 at point P(x, y).
By using the section formula, coordinates of P(x, y) are:
\( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here \( x_1 = -2, x_2 = 6, y_1 = -5, y_2 = 3, m = k, n = 1 \)
\( P(x, y) = \left( \frac{6k - 2}{k + 1}, \frac{3k - 5}{k + 1} \right) \)
But, P lies on x – 3y = 0

\( \implies x = 3y \)

\( \implies \frac{6k - 2}{k + 1} = 3\left( \frac{3k - 5}{k + 1} \right) \)

\( \implies 6k - 2 = 9k - 15 \)

\( \implies 3k = 13 \)

\( \implies k = \frac{13}{3} \)
Coordinates of P are:
\( \left( \frac{6 \times \frac{13}{3} - 2}{\frac{13}{3} + 1}, \frac{3 \times \frac{13}{3} - 5}{\frac{13}{3} + 1} \right) = \left( \frac{24 \times 3}{16}, \frac{8 \times 3}{16} \right) = \left( \frac{9}{2}, \frac{3}{2} \right) \)
Hence, the coordinates of point of intersection ‘P’ are \( \frac{9}{2} \) and \( \frac{3}{2} \).

Question. Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A(–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.
Answer: Let Q (x, 0) be the point on the x-axis which lies on the perpendicular bisector of AB.
\( \therefore QA = QB \implies (QA)^2 = (QB)^2 \)

\( \implies (-5 - x)^2 + (-2 - 0)^2 = (4 - x)^2 + (-2 - 0)^2 \)

\( \implies 25 + x^2 + 10x + 4 = 16 + x^2 - 8x + 4 \)

\( \implies 10x + 8x = 16 - 25 \)

\( \implies 18x = -9 \implies x = -\frac{1}{2} \)
Hence, the point Q is \( \left( -\frac{1}{2}, 0 \right) \).
Now \( QA^2 = \left( -5 + \frac{1}{2} \right)^2 + (-2 - 0)^2 = \left( -\frac{9}{2} \right)^2 + 4 = \frac{81}{4} + \frac{4}{1} = \frac{81 + 16}{4} = \frac{97}{4} \)

\( \implies QA = \frac{\sqrt{97}}{2} \) units
Now \( QB^2 = \left( 4 + \frac{1}{2} \right)^2 + (-2 - 0)^2 = \left( \frac{9}{2} \right)^2 + (-2)^2 = \frac{81}{4} + 4 = \frac{81 + 16}{4} = \frac{97}{4} \)

\( \implies QB = \frac{\sqrt{97}}{2} \) units
\( AB = \sqrt{(4 + 5)^2 + (-2 + 2)^2} = \sqrt{9^2} = 9 \) units

\( \implies AB = 9 \) units and \( QA = QB = \frac{\sqrt{97}}{2} \) units
Hence, \( \Delta QAB \) is an isosceles \( \Delta \).

Question. Find the point on y-axis which is equidistant from the points (5, – 2) and (– 3, 2).
Answer: Let the required point on y-axis be P(0, b).
Given, points are A(5, – 2) and B(– 3, 2).
Since the points A and B are equidistant from point P, the distance AP = distance BP
Applying the distance formula, we get:
\( \sqrt{(x_1 - x)^2 + (y_1 - y)^2} = \sqrt{(x_2 - x)^2 + (y_2 - y)^2} \)
Here, \( x = 0, y = b, x_1 = 5, y_1 = -2, x_2 = -3, y_2 = 2 \)

\( \implies \sqrt{(5 - 0)^2 + (-2 - b)^2} = \sqrt{(-3 - 0)^2 + (2 - b)^2} \)
On squaring both sides, we get

\( \implies 25 + (-2 - b)^2 = 9 + (2 - b)^2 \)

\( \implies 25 + 4 + b^2 + 4b = 9 + 4 + b^2 - 4b \)

\( \implies 8b = -16 \)

\( \implies b = -2 \)
Hence, the required point is (0, – 2).

Question. If the point A (2, –4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.
Answer: It is given that A (2, –4) is equidistant from P (3, 8) and Q (–10, y).
We know that distance between the points \( (x_1, y_1) \) and \( (x_2, y_2) \), \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( \implies \) Distance between P (3, 8) and A (2, –4) = Distance between A (2, –4) and Q (–10, y)

\( \implies \sqrt{(2-3)^2 + (-4-8)^2} = \sqrt{(-10-2)^2 + (y+4)^2} \)

\( \implies \sqrt{1^2 + 12^2} = \sqrt{12^2 + (y+4)^2} \)
Squaring both sides, we get
\( 1^2 + 12^2 = 12^2 + (y+4)^2 \)

\( \implies 1 + 144 = 144 + y^2 + 16 + 8y \)

\( \implies y^2 + 8y + 15 = 0 \)

\( \implies y^2 + 5y + 3y + 15 = 0 \)

\( \implies y(y+5) + 3(y+5) = 0 \)

\( \implies (y+5)(y+3) = 0 \)
If y + 5 = 0 then y = –5
If y + 3 = 0 then y = –3
Distance between P (3, 8) and Q (–10, y) when y = –3
\( PQ = \sqrt{(-10 - 3)^2 + (-3 - 8)^2} = \sqrt{(-13)^2 + (-11)^2} = \sqrt{169 + 121} = \sqrt{290} \)
Distance between P (3, 8) and Q (–10, y) when y = –5
\( PQ = \sqrt{(-10 - 3)^2 + (-5 - 8)^2} = \sqrt{(-13)^2 + (-13)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2} \)
Hence, the values of y are –3 and –5 and the corresponding values of PQ are \( \sqrt{290} \) and \( 13\sqrt{2} \).

Question. If A(–2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Also, find the lengths of its sides.
Answer: Given, ABCD is a parallelogram, in which diagonals AC and BD bisect each other at O. Coordinates of A(–2, 1) are given. Now, O is the mid-point of AC and BD, respectively.
By the mid point formula: Coordinates of \( O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \)
For line BD, \( O(x, y) = \left( \frac{a+1}{2}, \frac{0+2}{2} \right) = \left( \frac{a+1}{2}, 1 \right) \) ...(i)
For line AC, \( O(x, y) = \left( \frac{-2+4}{2}, \frac{1+b}{2} \right) = \left( 1, \frac{1+b}{2} \right) \) ...(ii)
From (i) and (ii), we get:
\( \frac{a+1}{2} = 1 \), \( 1 = \frac{1+b}{2} \)

\( \implies a = 1, b = 1 \)
Length of side AB, using the distance formula:
\( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(1+2)^2 + (0-1)^2} = \sqrt{9+1} = \sqrt{10} \) units
\( \therefore AB = CD = \sqrt{10} \) units. Similarly, all the four sides of parallelogram is \( \sqrt{10} \) units.

Question. Find the coordinates of the points of trisection of the line segment joining the points (3, – 2) and (–3, –4).
Answer: The given line segment is A(3, – 2) and (–3, –4).
Here, C(x, y) and C'(x', y') are the points of trisection of AB.
Then, AC : CB = 1 : 2 and AC' : C'B = 2 : 1.
By section formula, Coordinates of C(x, y) are \( \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \)
Here, m = 1, n = 2, \( x_1 = 3, y_1 = -2, x_2 = -3, y_2 = -4 \)
\( C(x, y) = \left( \frac{1 \times (-3) + 2 \times 3}{1+2}, \frac{1 \times (-4) + 2 \times (-2)}{1+2} \right) = \left( \frac{-3 + 6}{3}, \frac{-4 - 4}{3} \right) = (1, -8/3) \)
Now, coordinates of C' are \( = \left( \frac{m'x_2 + n'x_1}{m'+n'}, \frac{m'y_2 + n'y_1}{m'+n'} \right) \)
Here, m' = 2, n' = 1
\( C'(x', y') = \left( \frac{2 \times (-3) + 1 \times 3}{1+2}, \frac{2 \times (-4) + 1 \times (-2)}{1+2} \right) = \left( \frac{-6 + 3}{3}, \frac{-8 - 2}{3} \right) = (-1, -10/3) \)
Hence, the coordinates of the trisection are \( (1, -8/3) \) and \( (-1, -10/3) \).

Question. If P (9a – 2, –b) divides line segment joining A (3a + 1, –3) and B (8a, 5) in the ratio 3:1, find the values of a and b.
Answer: It is given that P (9a – 2, –b) divides line segment joining A (3a + 1, –3) and B (8a, 5) in the ratio 3:1.
By section formula, the coordinates of P are given as:
\( \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
\( = \left( \frac{3(8a) + 1(3a + 1)}{3 + 1}, \frac{3(5) + 1(-3)}{3 + 1} \right) \)

\( \implies (9a - 2, -b) = \left( \frac{24a + 3a + 1}{4}, \frac{15 - 3}{4} \right) \)

\( \implies 9a - 2 = \frac{27a + 1}{4} \) and \( -b = \frac{12}{4} \)

\( \implies 36a - 8 = 27a + 1 \) and \( -4b = 12 \)

\( \implies 9a = 9 \) and \( b = -3 \)

\( \implies a = 1 \) and \( b = -3 \)
Hence, the required values of a and b are 1 and –3.

Question. Find the ratio in which line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.
Answer: Let the line 2x + 3y – 5 = 0 divide the line segment joining the points A(8, –9) and B(2, 1) in the ratio m : 1 at point P.
\( \therefore \) Coordinates of P, using section formula
\( (x, y) = \left( \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right) \)
\( (x, y) = \left( \frac{2m + 8}{m + 1}, \frac{m - 9}{m + 1} \right) \)
It is given that P lies on 2x + 3y – 5 = 0

\( \implies 2\left( \frac{2m + 8}{m + 1} \right) + 3\left( \frac{m - 9}{m + 1} \right) - 5 = 0 \)

\( \implies 2(2m + 8) + 3(m - 9) - 5(m + 1) = 0 \)

\( \implies 4m + 16 + 3m - 27 - 5m - 5 = 0 \)

\( \implies 2m - 16 = 0 \)

\( \implies m = \frac{16}{2} = 8 \)

\( \implies m:1 = 8:1 \)
\( \therefore \) Coordinates of point \( P = \left( \frac{2(8) + 8}{8 + 1}, \frac{8 - 9}{8 + 1} \right) = \left( \frac{24}{9}, -\frac{1}{9} \right) = \left( \frac{8}{3}, -\frac{1}{9} \right) \)
Hence, the required point of division is \( \left( \frac{8}{3}, -\frac{1}{9} \right) \).

Question. Find the coordinates of a point on the x-axis which is equidistant from the points A(2, – 5) and B(– 2, 9).
Answer: Let, the coordinates of points on x-axis be P(x, 0).
[y-coordinate will be zero, because point is on x-axis]
Point A(2, – 5) and point B(– 2, 9) are equidistant from point P.
By distance formula, PA = PB
\( \sqrt{(x_1 - x)^2 + (y_1 - y)^2} = \sqrt{(x_2 - x)^2 + (y_2 - y)^2} \)
Here, \( x_1 = 2, y_1 = -5, x_2 = -2, y_2 = 9 \)

\( \implies \sqrt{(2 - x)^2 + (-5)^2} = \sqrt{(-2 - x)^2 + (9)^2} \)
On squaring both sides, we get
\( 4 + x^2 - 4x + 25 = 4 + x^2 + 4x + 81 \)

\( \implies -8x = 81 - 25 \)

\( \implies x = -7 \)
Hence, the coordinates of a point on x-axis is (– 7, 0).

Question. Write the coordinates of a point P on x-axis which is equidistant from the points A(–2, 0) and B(6, 0).
Answer: Let the co-ordinates of P be (x, 0).
Here, the given points are A(–2, 0) and B(6, 0)
According to the question, PA = PB or \( PA^2 = PB^2 \)
By the distance formula, we have \( (x_1 - x)^2 + (y_1 - y)^2 = (x_2 - x)^2 + (y_2 - y)^2 \)
Here \( x_1 = -2, y_1 = 0, x_2 = 6, y_2 = 0 \)
\( (-2 - x)^2 + (0 - 0)^2 = (6 - x)^2 + (0 - 0)^2 \)

\( \implies 4 + x^2 + 4x = 36 + x^2 - 12x \)

\( \implies 16x = 32 \) or x = 2
Hence, the co-ordinate of P area (2, 0).

Question. If the coordinates of points A and B are (–2, –2) and (2, –4) respectively, find the coordinates of P such that \( AP = \frac{3}{7} AB \), where P lies on the line segment AB.
Answer: \( AP = \frac{3}{7} AB \implies AP : PB = 3 : 4 \)
A(–2, –2) and B(2, –4)
\( x = \frac{6 - 8}{7} = -\frac{2}{7} \)
\( y = \frac{-12 - 8}{7} = -\frac{20}{7} \)
\( P\left( -\frac{2}{7}, -\frac{20}{7} \right) \)

Question. The point R divides the line segment AB, where A (– 4, 0) and B (0, 6) such that \( AR = \frac{3}{4} AB \). Find the coordinates of R.
Answer: Given, AB line segment with coordinates of A as (– 4, 0) and coordinates of B as (0, 6).
and \( AR = \frac{3}{4} AB \)

\( \implies 4AR = 3(AR + RB) \)

\( \implies AR = 3RB \)

\( \implies \frac{AR}{RB} = \frac{3}{1} \)
Let the coordinates of R be (x, y). By the section formula,
\( R(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here, m = 3, n = 1, \( x_1 = -4, y_1 = 0, x_2 = 0, y_2 = 6 \)
\( \therefore R(x, y) = \left( \frac{3 \times 0 + 1 \times (-4)}{3 + 1}, \frac{3 \times 6 + 1 \times 0}{3 + 1} \right) \)
\( \therefore R(x, y) = \left( -1, \frac{9}{2} \right) \)
Hence, the coordinates of R are: \( \left( -1, \frac{9}{2} \right) \).

Question. Find the ratio in which the line segment joining the points (1, – 3) and (4, 5) is divided by x-axis? Also find the coordinates of this point on x-axis.
Answer: Let, the required point be (a, 0) and the required ratio be k : 1.
By the section formula, the co-ordinates of P are given by:
\( P(x, y) = \left[ \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right] \)
Here, m = k, n = 1, \( x_1 = 1, x_2 = 4, y_1 = -3, y_2 = 5, x = a, y = 0 \)
\( a = \frac{4k + 1}{k + 1} \), and \( 0 = \frac{5k - 3}{k + 1} \)

\( \implies 5k - 3 = 0 \)

\( \implies k = \frac{3}{5} \)
Now, \( a = \frac{4(3/5) + 1}{(3/5) + 1} = \frac{12/5 + 1}{8/5} = \frac{17}{8} \)
Hence, the ratio in which the line segment is divided is 3 : 5 and the coordinates of the point are \( \left( \frac{17}{8}, 0 \right) \).

Question. Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, – 3). Hence find m.
Answer: Let the ratio k : 1.
Here, the coordinates of A are (2, 3) B are (6, – 3) and division point are (4, m).
By the section formula \( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here \( x_1 = 2, y_1 = 3, x_2 = 6, y_2 = -3, x = 4, y = m, m = k, n = 1 \)
\( P(4, m) = \left( \frac{k \times 6 + 1 \times 2}{k + 1}, \frac{k \times (-3) + 1 \times (3)}{k + 1} \right) \)
\( \therefore 4 = \frac{6k + 2}{k + 1} \)

\( \implies 4k + 4 = 6k + 2 \)

\( \implies 2k = 2 \)

\( \implies k = 1 \)
\( \therefore \) Ratio in which P divides AB is 1 : 1.
Now \( m = \frac{-3(1) + 3}{1 + 1} = 0 \)
Hence, the value of m is 0.

Question. If the line segment joining the points A(2, 1) and B(5, –8) is trisected at the point P and Q, find the coordinates of P.
Answer: Given, line segment A(2, 1) and B(5, –8) is trisected at point P and Q.
Then AP : PB = 1 : 2 and AQ : QB = 2 : 1
Let the coordinates of P be (x, y) and Q be (x’, y’)
Now, on applying the section formula, \( P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \)
Here, \( x_1 = 2, y_1 = 1, x_2 = 5, y_2 = -8, m = 1, n = 2 \)
\( \therefore P(x, y) = \left[ \frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2} \right] = [3, -2] \)
Now, coordinates Q(x’, y’)
\( Q(x', y') = \left[ \frac{2 \times 5 + 1 \times 2}{1 + 2}, \frac{2 \times (-8) + 1 \times 1}{1 + 2} \right] = (4, -5) \)
Hence, the coordinates of P are (3, –2) and Q are (4, –5)

Long Answer Type Questions

Question. In the given figure, \( \Delta ABC \) is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.
Answer: Given an equilateral triangle ABC of side 3 units. Coordinates of vertex A are (2, 0). Let the coordinates of B = (x, 0) and C = (x', y').
Then, using the distance formula, \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( 3 = \sqrt{(x - 2)^2 + (0 - 0)^2} \)
On squaring both sides we get \( 9 = (x - 2)^2 \)

\( \implies x^2 + 4 - 4x = 9 \)

\( \implies x^2 - 4x - 5 = 0 \)

\( \implies x^2 - 5x + x - 5 = 0 \)

\( \implies (x - 5)(x + 1) = 0 \)

\( \implies x = 5, -1 \)
But \( x \neq -1 \) (since it lies on positive x-axis). Coordinates of B are (5, 0).
Now: AC = BC (Sides of an equilateral triangle are equal)
\( \implies (x' - 2)^2 + (y' - 0)^2 = (x' - 5)^2 + (y' - 0)^2 \)

\( \implies x'^2 + 4 - 4x' + y'^2 = x'^2 + 25 - 10x' + y'^2 \)

\( \implies 6x' = 21 \implies x' = 7/2 \)
But AC = 3 (given)
\( \sqrt{(x' - 2)^2 + (y' - 0)^2} = 3 \)

\( \implies x'^2 + 4 - 4x' + y'^2 = 9 \)

\( \implies \frac{49}{4} + 4 - 4 \times \frac{7}{2} + y'^2 = 9 \)

\( \implies y'^2 = 9 - \frac{9}{4} = \frac{27}{4} \implies y' = \pm \frac{3\sqrt{3}}{2} \)
But, C lies in the first quadrant. \( y' = \frac{3\sqrt{3}}{2} \). Coordinates of C are: \( \left( \frac{7}{2}, \frac{3\sqrt{3}}{2} \right) \).

Question. Show that \( \Delta ABC \), where A(–2, 0), B(2, 0), C(0, 2) and \( \Delta PQR \) where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles.
Answer: Proof: In \( \Delta ABC \), find the length of sides AB, BC and CA by distance formula.
\( AB = \sqrt{(2 + 2)^2 + (0 - 0)^2} = \sqrt{4^2 + 0^2} = 4 \)
\( BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \)
\( CA = \sqrt{(0 + 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \)
Now, in \( \Delta PQR \), length of side PQ, QR and PR:
\( PQ = \sqrt{(4 + 4)^2 + (0 - 0)^2} = \sqrt{8^2} = 8 \)
\( QR = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{16 + 16} = 4\sqrt{2} \)
\( PR = \sqrt{(0 + 4)^2 + (4 - 0)^2} = \sqrt{16 + 16} = 4\sqrt{2} \)
For \( \Delta ABC \) and \( \Delta PQR \) to be similar, their sides should be proportional.
\( \frac{AB}{PQ} = \frac{4}{8} = \frac{1}{2} \), \( \frac{BC}{QR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2} \), \( \frac{CA}{PR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2} \)
So \( \frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR} \implies \Delta ABC \sim \Delta PQR \). Hence proved.